This project as currently described is badly flawed in a number of ways that are liable to lead to complete failure of the equipment. As it is stated that it is intended to be used in a life support system, equipment failures could result in fatalities.

It must be understood that advice given here is by way of comment and, even if followed, is in no way certain to be adequate to ensure that lives are not endangered.

The poster should seek competent professional assistance before implementing any system for this purpose. - RM

Added: Dude8604 has indicated that this system is for a respirator for his own use. That makes the concerns just as valid BUT the risks are up to him to judge.

I'm building a 12V UPS system to run some critical medical equipment without power interruption. The power noeds to switch over immediately so the equipment won't shut off. I have it finished except for one problem - the relay takes about 5 seconds to switch over. I've traced the problem to the 13.8V switched mode power supply, which is connected to the relay coil and the normally open terminal. The output of the power supply stays at a voltage high enough to keep the relay closed, but too low to keep the equipment running for those 5 seconds.

I have 2 1N914 diodes in series with the relay coil to reduce the voltage to the coil to closer to 12V coil spec. Do I just need to put in a Zener diode with the right voltage to keep the relay barely closed? This doesn't seem like the best way to do it since the power supply's output capacitor won't discharge for a long time.

Or is there a way I can rapidly discharge the power supply's output capacitor low enough to switch the relay without wasting a lot of power through a resistor in parallel?

Here's a link to: the power supply

the relay

On a related note, I'm considering switching from relays to transistors. To get a "normally closed" position, I think I'd have to use a depletion mode P FET (with Vth=0? does that exist?), but they're nearly impossible to find, especially for a decent price. Or a P JFET, but the ones I've seen can only handle a few mA, unless I use it to drive the gate of an enhancement mode FET, but I'd prefer a single transistor solution.


edit: Here's the circuit diagram. My question is about the relay towards the bottom right. The rest is for charging and might not be the best way, but that's a separate issue for now.Schematic

edit 2: Simplified version: simplified version

  • \$\begingroup\$ can you show us a really simple block diagram of the system, and the connection to the relay that you want to switch? Is the equipment run off mains power normally, and when that fails you want your system to take over and switch in your own 12V supply? \$\endgroup\$
    – KyranF
    Commented Dec 13, 2014 at 7:53
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    \$\begingroup\$ RadioShack doesn't explicitly disclaim fitness for life support purposes, so I guess you are good to go with the power supply.</sarcasm> \$\endgroup\$
    – jippie
    Commented Dec 13, 2014 at 10:15
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    \$\begingroup\$ Dude8604 - This IS the right place to come for help. BUT you must listen carefully. I about never point it out but in this case it's a good idea. Please note my "rep". I'm oldish and highly experienced (50+ years experience :-( ) and 'know what I am doing' with electronics and have so far come closer to killing myself than anyone else. | Your enthusiasm is commendable but there are enough things wrong with what you are proposing that could cause the system to fail that it would be extremely wise to start again with assistance from competent people. .... \$\endgroup\$
    – Russell McMahon
    Commented Dec 13, 2014 at 23:47
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    \$\begingroup\$ .... People here are willing and able to help but the starting point is NOT to patch up this circuit but instead to understand your requirement so as to be able to help suggest a suitable solution. What country are you in. What / who will the respirator be used for? If you are in the US this may be illegal and if legal would certainly expose you to the risk of severe legal penalties if this is for human life support. | It MAY be that there IS in fact good reason for you to be building such a system (war zone / developing country with no hope of a professional system .... . But ... \$\endgroup\$
    – Russell McMahon
    Commented Dec 13, 2014 at 23:52
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    \$\begingroup\$ ... we need to know such things in order to provide best help. What WhatRoughBeast wrote is technically fairly good and, as he said, it's not complete. | Advice you have received on charging may not treat the batteries at all well (too low rather than too high a charge rate). | Much more. Help us help you by telling us more about your requirement and situation, as above. \$\endgroup\$
    – Russell McMahon
    Commented Dec 13, 2014 at 23:55

2 Answers 2


Jesus wept. Moderators - please close this question. Poster is asking for help in building equipment which is likely to get people killed. Worse, folks are helping him.

That said, let me expand. I'm assuming that when you say "critical medical equipment" you mean exactly that. If you are exaggerating, then I'm overreacting. If you're not, then you really need to pay attention.

First, you don't know what you're doing. I'm sorry, this is not a form of disrespect, but you need to be aware of your limits. Under other circumstances this would be cause for a certain amount of hand-holding, but other circumstances don't include putting faulty medical equipment into practice. In the event of a failure, you are at best looking at the possibility of a major professional negligence lawsuit, and at worst criminal charges (negligent homicide). Grieving relatives can be remarkably vindictive.

The following list should not be considered a check list of what you need to address to finish your project. Please. I'm trying to indicate to you why you need to back off.

  1. Are you aware of the certification requirements for medical equipment? Are you going to get your system certified? If not, why not?

  2. Your delay is caused by the fact that, according to the data sheet, your relay dropout voltage is about 10% of nominal - that is, 1.2 volts, more or less. Once the relay operates (pulls in) it will remain in contact until the coil voltage drops below the dropout voltage. This is part of the data sheet, and is standard relay behavior.

  3. Putting diodes (zener or otherwise) into the coil circuit will reduce the coil voltage, alright, and cause the relay to drop out earlier. And when the coil voltage is reduced to 80% of nominal the relays won't operate, either. Why don't you realize this?

  4. A dropping diode (zener or otherwise) will dissipate exactly as much power as a resistor which drops the same voltage at the same current. Why don't you realize this?

  5. Your relay coils are rated for a maximum continuous coil voltage of 110% of nominal, that is 13.2 volts. Until you put in a dropping diode you were exceeding the coil rating. Why don't you realize this?

  6. As has been commented, after you charge one of your batteries you run the risk of burning the contacts on the charging relay. Why don't you realize this?

  7. You're using a non-medical-rated power supply. This alone will get you sued if there's a problem. Why don't you realize this?

  8. You're using the decay of your power supply to cause dropout of the relays. The fast way to do this is to directly monitor the power line input. Why don't you realize this?

  9. You don't show it, but you need to monitor the DC power supply and the batteries, and the relay operation in the event of power failure, and provide an annoying alarm that cannot be silenced. Have you done so? If not, why not?

  10. What provisions have you made for the effects of a lightning strike on the power lines?

  11. Your power supply uses banana jacks for output power. What is the long-term reliability of banana jack/plug connections? What will prevent careless contact from disconnecting it? Do you have any idea how hard it is to make things foolproof? Fools can be so very clever.

  12. If the power supply quietly fails, the batteries will take over - until they're discharged. See point 6. You cannot simply assume that someone will notice the failure condition before the equipment shuts down.

  13. If the output is accidentally shorted, the output capacitors in your DC power supply may well burn or weld the contacts in your relay. What provision have you made against this possibility?

I'm sorry, but I'm out of energy here. But I could go on at length. Please, you are out of your depth in a pool where lives depend on your doing thing right.

  • \$\begingroup\$ +1. Agree - except re closure. See following. He is better off staying here and being advised what he SHOULD do (whatever it is) than being kicked out and going elsewhere for "help", which may well fail him. | I have added a large text note to the top of the question. | I have flagged this for moderator attention. If you have not done so and have enough rep to do so please do so. I suggested that we do NOT close the question but instead try to convince him that he needs to do differently | NB: MAY NOT be in US or similar. MAY just perhaps be for eg animals. | Think about F6 & F7 :-(. \$\endgroup\$
    – Russell McMahon
    Commented Dec 13, 2014 at 23:37
  • \$\begingroup\$ You're probably right, although there are probably less antognistic ways of saying that. But that's beside the point. This was originally going to be a secondary power source for long term power outages. This is for my own personal use so what I'm thinking now is to get a medical grade UPS which would last at least long enough for a good margin of safety. Then the supply I'm building would kick in to keep the UPS fully charged for at least several days. That way I would have the safety features and alarm of the UPS but would still be able to supplement the battery life. Does this seem safe? \$\endgroup\$
    – dude8604
    Commented Dec 14, 2014 at 0:01
  • \$\begingroup\$ So would the second circuit diagram I posted work for that? Even if I have the medical UPS I'd still like it to switch over quickly, and alarm to indicate that the power is out and also alarm when the battery is low enough that I need to think about making other arrangements and/or start the generator. \$\endgroup\$
    – dude8604
    Commented Dec 14, 2014 at 0:07
  • \$\begingroup\$ Oh. Well, if this is just for your personal use, you obviously can't sue yourself, so I guess I overreacted. Some. All the problems are still there, but if you're willing to take the chances I can't argue that hard. You still need to do a lot of research. Best of luck. \$\endgroup\$ Commented Dec 14, 2014 at 0:51
  • \$\begingroup\$ The simplified version won't work. You need a separate charger for each battery. Except for the switching issues your first version had the right idea. You might try this: Use the charging circuit in edit 1. Connect each battery output to the load via a high-current (>15 amps) diode. Set the DC supply to slightly higher than the peak battery charging voltage and connect it to load via a diode as well. Now no relay is required. If the DC supply voltage drops the batteries will take up the slack. You'll lose some energy to diode drop, but Schottky diodes will do (.66 volts @20A: VFT2045BP-M3/4W) \$\endgroup\$ Commented Dec 14, 2014 at 1:06

Perhaps you could just connect the 13.8 volt supply in parallel with the battery? The system always "runs from the battery" but also recharges the battery always when mains is on. If the battery is a lead-acid battery then you should have no problems. 13.8 volts is a good voltage to charge the battery to, but if you're worried that the battery charger might actually drain the battery (normally this wouldn't happen), then you can also add the diode.

  • \$\begingroup\$ Is that really okay for the batteries? Wouldn't it charge too quickly and overload the power supply? They're 5 x 100Ah deep cycle batteries which are charged individually with relays on a timer (in hindsight maybe not the best design) and switch to being in parallel when the power is out. \$\endgroup\$
    – dude8604
    Commented Dec 13, 2014 at 9:00
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    \$\begingroup\$ It's a bit dangerous to not have the lead-acid batteries always connected in parallel. If you charge them separately, they will probably have different voltages when you do connect them together in parallel. And then, hundreds of amperes of current will flow from the strongest battery to anothers and sparks will fly in your relays. I would have the batteries always connected in parallel so that they always maintain the same charge level and use a power supply with current limit, or indeed a battery charger, as the power supply. \$\endgroup\$
    – PkP
    Commented Dec 13, 2014 at 9:22
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    \$\begingroup\$ Dude, you say that "It's for respiratory equipment" - I'd say that it's of paramount importance that you run the respiratory equipment directly from the battery. You don't really need those relays for anything and if they should fail... I don't even want to think about it. Plus as I explained in my previous comment, I think those fuses F1...F5 will blow the instant you switch all the relays on together. \$\endgroup\$
    – PkP
    Commented Dec 13, 2014 at 9:31
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    \$\begingroup\$ A "true" UPS has batteries permanently powering the output (respirator) with the main supply keeping the batteries "topped up". Think of an Ipad, running while plugged into a charger. When you lose power (unplug the Ipad) the batteries keep on going. You might use a relay to detect loss of mains, and give a warning beep (because hey, that backup UPS battery won't last forever). \$\endgroup\$ Commented Dec 13, 2014 at 9:31
  • \$\begingroup\$ Yes, a battery charger will be ok. To be sure, you can check the ratings on the battery charger if it says anything about a maximum size of the battery (it shouldn't) and there wouldn't be any harm in contacting the manufacturer of the battery charger about whether he recommends the use of the product in life critical equipment or not. \$\endgroup\$
    – PkP
    Commented Dec 13, 2014 at 9:36

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